Back to the on-screen lesson ·
Alpha, beta and gamma decay, balancing decay equations, Q-values from atomic masses, recoil sharing, and activity $A = \lambda N$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to balance decay equations, compute Q-values and decay energies, and find the activity of a sample.
From the last lesson you know that nuclei have binding energies set by their mass defects, and that the most stable lie near iron. From lesson 20 you know alpha particles escape by tunneling. This lesson describes how unstable nuclei change, what they emit, how much energy they release, and how to measure a sample's rate of decay.
| Term | What it means |
|---|---|
| Alpha decay | Emission of a helium nucleus, $^4_2$He: $A - 4$, $Z - 2$. |
| Beta-minus decay | A neutron becomes a proton, emitting an electron and an antineutrino: $Z + 1$. |
| Beta-plus decay | A proton becomes a neutron, emitting a positron and a neutrino: $Z - 1$. |
| Gamma decay | An excited nucleus emits a photon; $A$ and $Z$ are unchanged. |
| Q-value | The energy released, $931.5$ MeV per u of mass lost. |
| Activity | $A = \lambda N$, decays per second, in becquerels (Bq). |
| Decay constant | $\lambda$, the probability per second that a given nucleus decays. |
An unstable nucleus becomes more stable by one of a few processes, each conserving nucleon number and charge:
The energy released, the Q-value, is the mass lost times $c^2$: $Q = 931.5\,\Delta m$ MeV for $\Delta m$ in u. Each nucleus has a fixed probability $\lambda$ of decaying per second, so a sample of $N$ nuclei has activity
$$A = \lambda N,$$
measured in becquerels, one decay per second.
Another way: picture
Picture a crowd of people each holding a die, rolling once a second, and leaving when they roll a six. No one knows who will leave next, and no one's chance depends on how long they have stayed. But a crowd of thousands shrinks by a predictable fraction every second. Radioactive nuclei behave exactly like that, with $\lambda$ playing the role of the chance of a six.
Another way: steps
Heavy nuclei, beyond about lead, carry so many protons that their electrical repulsion outweighs part of the binding. Emitting an alpha particle, the tightly bound helium-4 nucleus, lowers both $A$ and $Z$ and releases a few MeV. Uranium-238 becomes thorium-234 with $Q = 4.27$ MeV; radium-226 becomes radon-222 with $4.87$ MeV.
Because momentum is conserved and the alpha is much lighter than the daughter, the alpha takes nearly all of $Q$, a fraction $(A - 4)/A$, and the daughter recoils with the rest. Every alpha from a given decay has the same energy, a sharp line, which is how alpha spectroscopy identifies isotopes. Alphas are stopped by a single sheet of ordinary paper or the outer layer of skin, but are dangerous if an alpha emitter is inhaled.
A nucleus with too many neutrons converts one into a proton, emitting an electron: carbon-14 becomes nitrogen-14, releasing $0.156$ MeV. With atomic masses, the Q-value is simply the parent's mass minus the daughter's, because the new electron is accounted for by the daughter atom's extra electron.
Unlike alpha particles, beta electrons emerge with a continuous spread of energies up to $Q$. In 1930 Wolfgang Pauli proposed that an unseen neutral particle carried off the rest, and Enrico Fermi named it the neutrino. Clyde Cowan and Frederick Reines detected neutrinos from a reactor at the Savannah River Plant in South Carolina in 1956, and Reines shared the 1995 Nobel Prize.
Nuclei with too many protons can convert one into a neutron, emitting a positron, the antimatter twin of the electron. Fluorine-18, made in hospital cyclotrons, does this; each positron soon meets an electron and both annihilate into two $0.511$ MeV gamma rays, the signal PET scanners detect.
After alpha or beta decay, the daughter is often left in an excited nuclear state, and it drops to its ground state by emitting a gamma ray, a photon of a sharply defined energy, typically hundreds of keV to a few MeV. Cobalt-60's gamma rays at $1.17$ and $1.33$ MeV are used to sterilize medical equipment and treat cancer.
Each nucleus has a probability $\lambda$ per second of decaying, independent of its age and of everything around it: temperature, pressure and chemistry make no measurable difference. A sample of $N$ nuclei therefore undergoes $\lambda N$ decays per second. The unit is the becquerel, one decay per second; the older unit, the curie, is $3.7 \times 10^{10}$ Bq, the activity of a gram of radium-226.
Because $\lambda = \ln 2/T_{1/2}$, short-lived isotopes are intensely active per gram and long-lived ones barely so. A gram of cobalt-60, half-life $5.3$ years, has an activity of $42$ TBq; a gram of uranium-238, half-life $4.5$ billion years, only $12$ kBq. The next lesson follows how activity falls with time.
Checking an answer. Mass numbers and charges must balance. $Q$ must be positive for a decay that happens. Alpha energies must be a few MeV, beta Q-values keV to a few MeV. Activities per gram must be larger for shorter half-lives.
Using atomic masses instead of nuclear masses is a bookkeeping choice that makes electrons cancel automatically in alpha and beta-minus decay. For beta-plus decay it leaves two extra electron masses on the parent's side, so $Q = 931.5(M_P - M_D) - 1.022$ MeV, a detail that matters for light nuclei.
The constant decay probability is an experimental fact, tested over billions of years in natural radioactivity and in the laboratory. It is a quantum effect: a nucleus has no memory and no aging, because a quantum state that has not yet decayed is identical to a freshly formed one.
Everyone is exposed to natural radioactivity: potassium-40 in the body, radon gas from the ground, cosmic rays and carbon-14. The average American receives about $6$ millisieverts a year, half from natural sources and half from medical procedures. Radon, an alpha emitter from the decay of uranium in soil, is the largest natural source and the second leading cause of lung cancer in the United States.
The Environmental Protection Agency recommends testing homes for radon and fixing those above $148$ Bq per cubic meter of air, a level where about $150$ radon nuclei decay in each cubic meter every second. Simple venting systems reduce it by drawing soil gas away from the foundation.
Many decays lead to daughters that are themselves unstable. Uranium-238 decays through a chain of fourteen steps, eight alpha and six beta, passing through radium and radon, before reaching stable lead-206. Each step has its own half-life, from microseconds to billions of years.
In an old uranium ore, the chain reaches a steady balance in which every member decays at the same rate as uranium itself. That is why uranium ores contain radium and radon, as Marie and Pierre Curie discovered by chemically separating tons of pitchblende. It is also how geologists date rocks, from the ratio of lead to uranium.
Plotting every known nucleus by its numbers of protons and neutrons gives a narrow band of stable ones, the valley of stability. Light stable nuclei have about equal numbers; heavy ones need more neutrons than protons to dilute the electrical repulsion, reaching about $1.5$ neutrons per proton in lead.
Nuclei off the band decay toward it: those with excess neutrons by beta-minus, those with excess protons by beta-plus or electron capture, and the heaviest by alpha. Researchers at Michigan State's Facility for Rare Isotope Beams make nuclei far from the band to learn where it ends, the limits beyond which no nucleus can hold together even for an instant.
Most ionization smoke detectors sold in American hardware stores contain about $33$ kBq of americium-241, a fraction of a microgram. Its alpha particles, $5.5$ MeV each, ionize the air between two charged plates, letting a tiny current flow. When smoke drifts in, its particles soak up the ions, the current drops, and the alarm sounds.
Americium-241's half-life of $432$ years keeps the source steady for the detector's life, and its alphas cannot penetrate the detector's case or even skin. The Nuclear Regulatory Commission permits the devices in homes because the dose they give is far below that from natural background, while they save thousands of lives a year.
A quick calculation shows how little americium is needed. With $\lambda = 5.08 \times 10^{-11}$ per second, an activity of $33$ kBq takes about $6.5 \times 10^{14}$ nuclei, roughly a quarter of a microgram. When a detector reaches the end of its ten-year service life, the manufacturer's instructions usually allow it to be returned by mail for recycling of the source.
In 1956 Clyde Cowan and Frederick Reines placed tanks of water and cadmium beside a nuclear reactor at the Savannah River Plant in South Carolina. The reactor's beta decays produced a vast flux of antineutrinos, about $10^{13}$ per square centimeter each second. Rarely, one struck a proton in the water, making a neutron and a positron.
The positron's annihilation gamma rays, followed microseconds later by the gamma of the neutron's capture in cadmium, gave a unique double signal. A few events an hour confirmed Pauli's particle, twenty-six years after he proposed it to save energy conservation in beta decay. Reines received the Nobel Prize in 1995.
It is natural to think a radioactive nucleus decays because it has grown old or used up something inside. In fact every nucleus of a given kind has the same chance of decaying in the next second, whether it formed a moment ago or billions of years ago. A nucleus has no memory; only whole samples show predictable behavior.
A related error is to think radiation makes things radioactive. Alpha, beta and gamma rays deposit energy and can damage cells, but they do not make the materials they strike radioactive; only neutrons, absorbed by nuclei, commonly do that.
Write the alpha decay of radium-226, $Z = 88$.
$^{226}_{88}\text{Ra} \to\, ^{222}_{86}\text{Rn} + {}^4_2\text{He}$
Balance $A$ and $Z$.
Find the mass lost using atomic masses.
$226.025410 - 222.017578 - 4.002603 = 0.005229\ \text{u}$
Parent minus products.
Convert to the Q-value.
$Q = 0.005229 \times 931.5 = 4.87\ \text{MeV}$
Energy released.
Find the alpha's share.
$K_\alpha = 4.87 \times \dfrac{222}{226} = 4.78\ \text{MeV}$
Momentum conservation.
Find the radon's recoil.
$K_{\text{Rn}} = 0.09\ \text{MeV}$
The remainder.
Write the beta-minus decay of carbon-14.
$^{14}_6\text{C} \to\, ^{14}_7\text{N} + e^- + \bar{\nu}$
A neutron becomes a proton.
Find the mass difference of the atoms.
$14.003242 - 14.003074 = 0.000168\ \text{u}$
Parent minus daughter.
Convert to the Q-value.
$Q = 0.000168 \times 931.5 = 0.156\ \text{MeV}$
Energy released.
Find the electron's largest energy.
$K_{\max} \approx 0.156\ \text{MeV}$
When the antineutrino takes little.
Estimate the typical electron energy.
$\text{about } 0.05\ \text{MeV}$
Roughly a third of $Q$ on average.
Explain why the energies vary.
$\text{three bodies share the energy}$
Evidence for the neutrino.
A hospital source holds $2.0$ mg of cobalt-60, half-life $5.27$ years. Count the nuclei.
$N = \dfrac{2.0 \times 10^{-3}}{60} \times 6.022 \times 10^{23} = 2.0 \times 10^{19}$
Grams over molar mass times $N_A$.
Convert the half-life to seconds.
$5.27 \times 3.156 \times 10^7 = 1.66 \times 10^8\ \text{s}$
SI units.
Find the decay constant.
$\lambda = \dfrac{0.693}{1.66 \times 10^8} = 4.17 \times 10^{-9}\ \text{s}^{-1}$
Per second.
Find the activity.
$A = 4.17 \times 10^{-9} \times 2.0 \times 10^{19} = 8.4 \times 10^{10}\ \text{Bq}$
Decays per second.
Convert to curies.
$\dfrac{8.4 \times 10^{10}}{3.7 \times 10^{10}} = 2.3\ \text{Ci}$
The older unit.
Find the gamma rays per second.
$2 \times 8.4 \times 10^{10} = 1.7 \times 10^{11}$
Two per decay.
Find the gamma power.
$1.7 \times 10^{11} \times 1.25\ \text{MeV} \times 1.6 \times 10^{-13} = 0.034\ \text{W}$
Small, but lethal up close.
Subtract four from the mass number.
$A' = 210 - 4 = 206$
Nucleons conserved.
Subtract two from the atomic number.
$Z' = 84 - 2 = 82$
Charge conserved.
Name the daughter.
A nucleus of uranium-238, with $Z = 92$ and $A = 238$, emits an alpha particle. What are the atomic number and mass number of the daughter nucleus?
Complete the worked solution: cesium-137 beta-minus decays to barium-137. With atomic masses, the parent is heavier than the daughter by $0.001262$ u. Find the daughter's atomic number, the energy released in MeV, and the most kinetic energy the electron can carry in MeV, taking $1$ u $= 931.5$ MeV/$c^2$.
Raise the atomic number by one.
$Z' = Z + 1 =$ y
A neutron became a proton.
Convert the mass difference to energy.
$Q = 931.5(M_{\text{parent}} - M_{\text{daughter}}) =$ q
Atomic masses include the new electron.
Give the electron the most it can take.
$K_{e,\max} \approx Q =$ k
When the antineutrino gets almost nothing.
Explain the spread of electron energies.
$\text{shared with an antineutrino}$
Why beta spectra are continuous.
Match each kind of decay to how it changes the nucleus.
| A − 4, Z − 2 | A unchanged, Z + 1 | A unchanged, Z − 1 | A and Z unchanged | |
|---|---|---|---|---|
| alpha | ||||
| beta-minus | ||||
| beta-plus | ||||
| gamma |
polonium-210 alpha decays to lead. The parent's atomic mass exceeds the daughter's plus helium-4's by $0.005806$ u. With $1$ u $= 931.5$ MeV/$c^2$, fill in that mass difference in u, the Q-value in MeV, and the alpha particle's kinetic energy in MeV, using $K_\alpha = Q(A - 4)/A$ with $A = 210$.
| value | |
|---|---|
| mass lost (u) | |
| Q-value (MeV) | |
| alpha kinetic energy (MeV) |
An alpha decay releases $Q = 4.871$ MeV, shared between the alpha particle and a daughter nucleus of mass number $A$. Treating masses as proportional to mass numbers, write the daughter's recoil kinetic energy, in MeV, as a formula in $A$ (for $A$ much greater than 4).
Answer:
What is the activity of one gram of cesium-137, whose half-life is 30.1 years? Use $N_A = 6.022 \times 10^{23}$ per mole and $1$ year $= 3.156 \times 10^7$ s, and give the answer in GBq.
Answer: GBq
An ionization smoke detector sold in American hardware stores contains americium-241, half-life $432.2$ years, with an activity of $25.9$ kBq. With $1$ year $= 3.156 \times 10^7$ s and $1$ u $= 1.6605 \times 10^{-27}$ kg, how much americium does it contain, in nanograms?
Answer: ng
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An alpha decay releases $Q = 5.637$ MeV, shared between the alpha particle and a daughter nucleus of mass number $A$. Treating masses as proportional to mass numbers, write the daughter's recoil kinetic energy, in MeV, as a formula in $A$ (for $A$ much greater than 4).
Answer:
You can analyze radioactive decays. Explain to someone why the alpha particle takes almost all the energy of an alpha decay.
20. Your turn: polonium-210, $Z = 84$, alpha decays. Name the daughter's $Z$ and $A$., step 3
$^{206}_{82}\text{Pb}$
Stable lead-206.